Problem: Graph this system of equations and solve. $-2x-y = 3$ $3x-y = -2$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Solution: Convert the first equation, $-2x-y = 3$ , to slope-intercept form. $y = -2 x - 3$ The y-intercept for the first equation is $-3$ , so the first line must pass through the point $(0, -3)$ The slope for the first equation is $-2$ . Remember that the slope tells you rise over run. So in this case for every $2$ positions you move down (because it's negative) $1$ position to the right. $2$ positions down from $(0, -3)$ is $(1, -5)$ Graph the blue line so it passes through $(0, -3)$ and $(1, -5)$ Convert the second equation, $3x-y = -2$ , to slope-intercept form. $y = 3 x + 2$ The y-intercept for the second equation is $2$ , so the second line must pass through the point $(0, 2)$ The slope for the second equation is $3$ . Remember that the slope tells you rise over run. So in this case for every $3$ positions you move up $1$ position to the right. $3$ positions up from $(0, 2)$ is $(1, 5)$ Graph the green line so it passes through $(0, 2)$ and $(1, 5)$ The solution is the point where the two lines intersect. The lines intersect at $(-1, -1)$.